/sci/ - Science & Math

Thanks for posting the version with 0% as an option.

this >>5724098 was meant to be for you >>5724015

fuck

>>5724008

10^5 - 7^5

Go to youtube and check:
/watch?v=tb9jwxr0HyU

It's a basic explanation of complimentary counting. Good luck, OP.

B

>>5724121
>10^5 - 7^5
That's for 3 OR 5 OR 7 appearing in the string -- not and.

>>5724099
how is it wrong, is it actually A?

is it 6000?

>>5724446
>he's never heard of a paradox

>>5724008
>>5724455
>>5724411

there is no paradox or recursion

it's a shittily worded question. If you assume there's only 1 right answer out of the 4, then you're going to have a 1/4 chance of getting it right.

the number %'s of each choice have nothing to do with whether or not the answer is right

>>5724456
That's what I got, C(3,5)*(10^2*3!).

>>5724458
>the number %'s of each choice have nothing to do with whether or not the answer is right
i'm a retard, please explain

Once again, /sci/ shows their lack of talent at avoiding trolls.

>>5724460
Oh, should be C(5,3)*(10^2*3!).

>>5724456
>>5724466
I'm getting
<span class="math">
\binom{5}{3} \times 10^{5-3}
[/spoiler]

>>5724458
its not worded poorly, 'an answer' implies any of the below could be an answer or any amount could be an answer. from then we have only the answers given to surmise the rest.

>>5724456
>>5724466
I'm getting
<span class="math">
{5 \choose 3} \times 10^{5-3}
[/spoiler]

>>5724459

the question should be split up in to two parts to remove ambiguity

there's no way to know what's actually being asked here

>>5724466
I got P(5,3)*(10^2)

>>5724476
WRONG

>>5724493
is this it?

(100 (Integrate[t^5/E^t, {t, 1, Infinity}] + Sum[(-1)^k/((6 + k) k!), {k, 0, Infinity}]))/(Integrate[t^2/E^t, {t, 1, Infinity}] + Sum[(-1)^k/((3 + k) k!), {k, 0, Infinity}])

>>5724493
Why don't you fucking explain it instead of being an autistic faggot.

For any particular string containing 3, 5 and 7, there will be two "free" spaces left, which means we simply need to find the number of ways to have 3, 5 and 7 in a five-digit string, then multiply that by 100 (since the remaining two digits in any such string can each be anything from 0 to 9).

Since order matters, we should use a permutation not a combination, so we have P(5,3)*100 = 6000, as seen in >>5724490

That said, C(5,3)(100*3!) still works, since C(5,3) = 5!/(3!2!) while P(5,3) = 5!/2!

>>5724543
Couldn't we be counting some strings twice with this though?

Like, if x can be anything from 0 to 9, and we have
xx357, x3x57 and 3xx57, then wouldn't the string 33357 be in each? Am I thinking of this incorrectly?

>>5724543
To continue because I'm exhausted and in a rambling mood, we can also do this by a bit of direct counting and multiplication.

For 3, 5 and 7, there are six possible orders: 357, 375, 537, 573, 735, 753.

For each of these, there are ten possible positions in the string. Taking 357 as an example, the ten possible positions are:

357xx, 35x7x, 35xx7, 3x57x, 3x5x7, 3xx57, x357x, x35x7, x3x57, xx357

Thus we see there are ten possible positions for each of our six possible orders, so 60 total, multiplied by the 100 possibilities for the remaining two digits in any valid string, thus 60*100 = 6000 valid strings.

>>5724568
Hm, that's a good point.

>>5724569
i got it here
>>5724504

>>5724569
or more easily:
you can put 1st digit in 5 different places, 2nd in remaining 4 and 3rd in remaining 3 => 5*4*3=60

>>5724587
True, but I think >>5724568 has established I was wrong anyway.

I'm getting 4350 using this code in Java:
http://pastebin.com/Grz8QZAX

>>5724568
I think we could do it as follows:

There are 10^5 total possible strings.

There are 9^5 that contain no 3's.

For 5, we have to find the number of strings that contain no 5 but do contain at least one 3 (since the ones with no 5 and no 3 were counted earlier). Thus we subtract the number of strings that contain no 5 and no 3 from the number of strings that simply contain no 5. This number is 9^5 - 8^5.

For 7, we find the number of strings that contain no 7, 9^5. From this we have to subtract the number of strings with no 7, no 3, and at least one 5 (8^5 - 7^5); no 7, no 5, and at least one 3 (8^5 - 7^5); and no 7, no 5, and no 3 (7^5).

Thus I arrive at 10^5 - (9^5) - (9^5 - 8^5) - (9^5 - 2(8^5 - 7^5) - 7^5) = 4350 valid strings.

I may have missed something, but I'm posting this anyway because I'm exhausted and therefore someone else is more likely to notice what (if anything) I missed and figure out a better solution.

>>5724615
Based on this, I think my methods works, then.

>>5724615
Based on this, I think my method in >>5724650 works, then. Thank you.

>>5724008
>>5724008
>>5724008
>>5724008
Ok you fucking niggers, if you're so smart, tell me the correct answer, and I'll tell you why it's wrong.

>>5725510
Actually, screw that, I'll tell you right now

>The answer is %0
No, because if the answer was %0, that would mean that you couldn't get the correct answer, and that there is no correct answer, when you just claimed that %0 WAS the correct answer

>25%
If %25 was the correct answer, you'd have a %50 chance if getting it, since there are two of them
>%50
There isn't a %50 chance of getting %50, there's only a %25 chance of getting %50, since there's only one %50

Jesus, you're THAT fucking kid from school who thought questions like "Can God make a rock so big he can't lift it?" were intelligent and cool, when in reality he was a fucking moron just like you.

>>5725510
>>5725517

imagine your face when you realized that wasn't even the point of the thread